Are M and N conjugate of each other $M = [ \begin{smallmatrix} x & 0 \\ 0 & y \end{smallmatrix}]$ and $N =[ \begin{smallmatrix} x & 1 \\ 0 & y \end{smallmatrix}]$, $x \neq y$ are conjugate for all $x,y$ in real numbers.
I want a hint to start with ! how can I find $G$ such that $GMG^{-1}=N$
 A: The eigenvalues of a triangular matrix are the entries in the diagonal; since $x\ne y$, the second matrix is diagonalizable.
The eigenvalues are the same, so $M$ and $N$ are similar, that is, conjugate.

An eigenvector for the eigenvalue $x$ is a nonzero vector in the null space of $N-xI$:
$$
N-xI=\begin{bmatrix} 0 & 1 \\ 0 & y-x \end{bmatrix}
$$
so, for instance, we can take
\begin{bmatrix}
1 \\ 0 
\end{bmatrix}
An eigenvalue for $y$ is a nonzero vector in the null space of $N-yI$:
$$
N-yI=\begin{bmatrix} x-y & 1 \\ 0 & 0 \end{bmatrix}
$$
so you can take
\begin{bmatrix} 1 \\ y-x \end{bmatrix}
Take
$$
G=\begin{bmatrix} 1 & 1 \\ 0 & y-x \end{bmatrix}
$$
and you have
$$
N=GMG^{-1}
$$
by standard facts on diagonalization. Try and perform the computation, but it's not really needed.
A: The matrix $M$ has eigenvalues $x$ and $y$ and the obvious eigenvectors are $e_1$ and $e_2$, resepectively. The matrix $N$ has the same characteristic polynomial and hence the same eigenvalues. For the first eigenvalue $x$ we still have $N e_1 = x e_1$. For the second eigenvalue $y$ we compute the eigenspace as
$$
\ker(N-yE_2) = 
\ker
\begin{bmatrix}
x-y & 1 \\
0 & 0
\end{bmatrix}
= \left\langle
\begin{pmatrix}
1 \\
y-x
\end{pmatrix}
\right\rangle.
$$
Hence you need to perform a base change from $e_1, e_2$ to $e_1, e_1+(y-x)e_2$.
Can you write down $G$ or $G^{-1}$ from this information?
A: Hint: You can take $\pmatrix{1\\0}$ as the first column of $G$, and rearrange as $GM=NG$. 
